a) In how many ways the letters of the word CALCULUS can be arranged so that vowels are always together? [2] b) Find the middle term in the expansion of (x 1x^2 )^12 . [2] c) Using Caley's table for…

a) Arrangements of "CALCULUS" with vowels together

Step 1: Identify vowels and consonants The word is CALCULUS.

• Vowels: A, U, U (3 vowels)
• Consonants: C, L, C, L, S (5 consonants)

Step 2: Treat vowels as a single entity Since vowels must stay together, we group them as one "block":

• Vowel block: (A, U, U)
• Consonants: C, L, C, L, S

Total entities to arrange = 1 (vowel block) + 5 (consonants) = 6 entities.

Step 3: Calculate arrangements of consonants + vowel block The consonants and vowel block can be arranged in: $$6!=720\text{ ways}$$

Step 4: Arrange vowels within their block The vowels A, U, U can be arranged among themselves. Since there are repeated U’s, the number of distinct arrangements is: $$\frac{3!}{2!}=\frac{6}{2}=3\text{ ways}$$

Step 5: Total arrangements Multiply the arrangements of the blocks by the arrangements within the vowel block: $$6!\times \frac{3!}{2!}=720\times 3=2160$$

Final answer: $$\boxed{2160}$$

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b) Middle term in the expansion of ${\left(x-\frac{1}{x^2}\right)}^{12}$

Step 1: Determine the number of terms The expansion of $(a+b{)}^n$ has $n+1$ terms. Here, $n=12$, so there are 13 terms.

Step 2: Identify the middle term Since 13 is odd, the middle term is the $\frac{13+1}{2}=7^{th}$ term.

Step 3: Use the binomial expansion formula The general term in the expansion of $(a+b{)}^n$ is: $$T_{k+1}=\left(\genfrac{}{}{0ex}{}{n}{k}\right)a^{n-k}{b}^k$$ Here, $a=x$, $b=-\frac{1}{x^2}$, and $n=12$. For the $7^{th}$ term ($k=6$): $$T_7=\left(\genfrac{}{}{0ex}{}{12}{6}\right)x^{12-6}{\left(-\frac{1}{x^2}\right)}^6$$

Step 4: Simplify the expression $$T_7=\left(\genfrac{}{}{0ex}{}{12}{6}\right)x^6{\left(-\frac{1}{x^2}\right)}^6=\left(\genfrac{}{}{0ex}{}{12}{6}\right)x^6\cdot \frac{1}{x^{12}}=\left(\genfrac{}{}{0ex}{}{12}{6}\right)x^{-6}$$ $$\left(\genfrac{}{}{0ex}{}{12}{6}\right)=924$$ So, $$T_7=924\cdot x^{-6}=\frac{924}{x^6}$$

Final answer: $$\boxed{\frac{924}{x^6}}$$

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c) Inverse elements in Cayley’s table for $(G,\ast )$ where $G=\{-1,1\}$

Step 1: Construct the Cayley’s table The group $G=\{-1,1\}$ with operation $\ast$ (multiplication) has the following table:

Explanation:

• $(-1)\ast (-1)=1$
• $(-1)\ast 1=-1$
• $1\ast (-1)=-1$
• $1\ast 1=1$

Step 2: Find inverses An element $a\in G$ has an inverse $b\in G$ if $a\ast b=e$, where $e$ is the identity element.

From the table, the identity element $e=1$.

• For $1$: $1\ast 1=1$, so the inverse of $1$ is $1$.

• For $-1$: $(-1)\ast (-1)=1$, so the inverse of $-1$ is $-1$.

Final answer: The inverse elements of $G$ are: $$\boxed{\begin{array}{rl}& \text{Inverse of }1\text{ is }1,\\ & \text{Inverse of }-1\text{ is }-1.\end{array}}$$